$A$ and $B$ connected implies at least one of $A \cup B$ or $A \cap B$ is connected

connectednessgeneral-topology

Let $A$ and $B$ be connected subsets of a metric space. Prove that at
least one of $A \cup B$ or $A \cap B$ is connected

My two attempts:

  1. Assume neither $A \cup B$ nor $A\cap B$ are connected (hence they are disconnected, then we can partition them $$A\cup B \rightarrow \{U_1,U_2\}\quad A \cap B \rightarrow \{V_1,V_2\} $$
    for some open, disjoint sets $U_1,U_2,V_1,V_2$. From here I'm not sure where to continue, and I suppose I need to arrive to a contradiction showing that either $A$ or $B$ is disconnected, which can't be as they are given to be connected but I'm not sure how to reach there.
  2. I tried to go about it by showing that if one of $A \cup B$ or $A \cap B$ is disconnected the other one must be connceted. Let $A \cup B$ be disconnected, then it can be partionted into two open disjoint sets $\{U_1,U_2\}$. Now fix $x\in A \cap B$ and then see that $x \in U_1$ or $x \in U_2$. WLOG choose the former. So we have $x \in A$ and $x\in B$ and $x \in U_1$. But once again I'm not sure how to use this to arrive to the fact to show that $A \cap B$ is connected

Any help would be appreciated.

Best Answer

There are two possibilities:

  1. $A\cap B=\emptyset$: then $A\cap B$ is connected.
  2. $A\cap B\neq\emptyset$: then $A\cup B$ is connected, because if $f\colon A\cup B\longrightarrow\{0,1\}$ (with $\{0,1\}$ endowed with the discrete topology) is continuous then, if $p\in A\cap B$, $f(A)=\bigl\{f(p)\bigr\}=f(B)$, and therefore $f$ is constant.
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